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Also 54—24=30 first diff. 54—36=18 second diff and 54–48=6 third diff.

Whence 54X 30 X 18 X 6 ✓174960 = 418.282 = area required.

2. Required the area of a triangle whose three sides are 13, 14, and 15 feet.

Ans. 84 square feet. 3. How many acres are there in a triangle whose three sides are 49.00, 50.25 and 25.69 chains? Ans. 61.498 ac.

4. Required the area of a right angled triangle, whose hypothenuse is 50, and the other two sides 30 and 40.

Ans. 600. 5. Required the area of an equilateral triangle, whose side is 25.

Ans. 270.6329. 6. Required the area of an isosceles triangle, whose base is 20, and each of its equal sides 15. Ans. 111.803.

7. Required the area of a triangle, whose three sides are 20, 30, and 40 chains.

Ans. 29 ac.7.58 po.

PROBLEM IV.

Any two sides of a right angled triangle being given to

find the third side.

RULE.*

. 1. When the two legs are given to find the hypothenuse

Add the square of one of the legs to the square of the other, and the square root of the sum will be equal to the hypothenuse. 2. When the hypothenuse and one of the legs are given

to find the other leg. From the square of the hypothenuse take the square of ahe given leg, and the square root of the remainder will be equal to the other leg.

* By Euc. 47.I.AB? + BC2 = ACP, or AC2-ABP =BC? ; and herefore V AB? + BC2 = AC, or V AC2

- AB? = BC, or Vac2-BC2=AB, which is the same as the rule.

LXAMPLES.

1. In the right angled triangle ABC, the base AB is 56 nd the perpendicular BC 33; what is the hypothenuse?

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Here 562 + 332 =3136+1089=4225; and / 4225=65 =hypothenuse AC.

2. If the hypothenuse AC be 53, and the base AB 45 ; what is the perpendicular BC?

Here 532—452=2809–2025=784; and ✓784=28= perpendicular BC.

3. The base of a right angled triangle is 77, and the perpendicular 36 : what is the hypothenuse? Ans. 85.

4. The hypothenuse of a right angled triangle is 109, and the perpendicular 60 : what is the base ? Ans. 91.

5. It is required to find the length of a shore, which strutting 12 feet from the upright of a building, will support a jamb 20 feet from the ground. Ans. 23.3238 feet.

6. The height of a precipice, standing close by the side of a river, is 103 feet, and a line of 320 feet will reach from the top of it to the opposite bank : required the breadtn of the river.

Ans. 302.9703 feet 7. A ladder 50 feet long, being placed in a street, reached window 28 feet from the ground, on one side ; and by turn ing it over, without removing the foot, it reached another window, 36 feet high, on the other side; required the breadth of the street.

Ans. 76.1233 feet

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PROBLEM V.

To find the area of a trapezium.

RULE.* Multiply the diagonal by the sum of the two perpendicu lars falling upon it froin the opposite angles, and half the product will be the area.

EXAMPLES.

1. Required the area of the trapezium BAED, whose diagonal BE is 84, the perpendicular AC 21, and DF 28.

D

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4116 Here 28+21 X 84=49 x 84=4116 ; and =2058

2 the area required.

BE X DF * Demon. The area of the triangle BDE is

;

and

2

BE XAC the area of the triangle BAE is= ; and therefore the

2 sum of these areas, or the area of the whole trapezium,

BE X DF BEX AC DF+AC is

X BF.

Q. E. D. 2 2

2 If the trapezium can be inscribed in a circle; that is, if the sum of two of its opposite angles is equal to two right angles, or 180°, the area may be found thus :

Rule. From half the sum of the four sides subtract each side seve. rally; then multiply the four remainders continually together, and the square root of the product will be the area.

+

=

2. Required the area of a trapezium whose diagonal is 80.5, and the two perpendiculars 24.5 and 30.1.

Ans. 2197.65. 3. What is the area of a trapezium whose diagonal is 108 feet 6 inches, and the perpendiculars 56 feet 3 inches, and 60 feet 9 inches ?

Ans. 6347 fe. 3 in

PROBLEM VI.

To find the area of a trapezoid, or a quadrangle, two of

whose opposite sides are parallel.

RULE.*

Multiply the sum of the parallel sides by the perpendicu. lar distance between them, and half the product will be the

area.

EXAMPLES.

1. Required the area of the trapezoid ABCD, whose sides AB and DC are 321.51 and 214.24, and perpendicular DE 171.16.

D C

n

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AB X DE * Demon. The A ABD is =

, and the A BCD=

2 OG X Bn

DCX DE or, (because on=DE)

Therefore, a 2

2

AB X D ABD X ABCD, or the whole trapezoid ABCD, is =

2 DCX DE AB+DC Х 2

2

X DE. Q. E. D.

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Here 321.51 +214.245535.75=sum of the parallel sides AB, DC. Whence 535.75X171.16 (the perp. DE)=91698.9700. And 91698.9700

=45849.485 the area required. 2. 2 The parallel sides of a trapezoid are 12.11 and 8.22 chains, and the perpendicular distance 5.15 chains; required the area

ac. 10. po.

Ans. 5 1 9.956 3. Required the area of a trapezoid whose parallel sides are 25 feet 6 inches and 18 feet 9 inches, and the perpendicular distance 10 feet 5 inches.

fe. in. pa.

Ans. 230 57 4. Required the area of a trapezoid whose parallel sides are 20.5 and 12.25, and perpendicular distance 10.75.

Ans. 176.03125

PROBLEM VII.

To find the area of a regular polygon.

RULE.*

Multiply half the perimeter of the figure by the perpen. dicular falling from its centre upon one of the sides, and the product will be the area.

Note. The perimeter of any figure is the sum of all its sides.

* Demon. Every regular polygon is composed of as many equal triangles as it has sides, consequently the area of one of those tri. angles being multiplied by the number of sides must give the area of the whole figure; but the area of either of the triangles is equal to the rectangle of the perpendicular and half the base, and therefore the rectangle of the perpendicular and half the sum of the sides is equal to the area of the whole polygon; thus,

5AB OPX-=area of the A AOR, and OP X-=area of the

polygon ARCDE. Q. E. D.

AB

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